Question

In Young's double slit experiment, if the distance between the slits is 2 mm and the distance of the screen from the slits is 100 cm, the fringe width is 0.36 mm. If the distance between the slits is decreased by 0.5 mm and the distance of the screen from the slits is increased by 50 cm, the fringe width becomes

• A. 0.84 mm
• B. 0.96 mm
• C. 0.48 mm
• D. 0.72 mm

Hint:

Fringe width varies directly with screen distance and inversely with slit separation in Young's double slit experiment.

Answer 1

The Correct Option is D

Fringe width \(\beta\) is given by: \[ \beta = \frac{\lambda D}{d} \] Given: Initial fringe width \(\beta_1 = 0.36\, mm\), slit distance \(d_1 = 2\, mm\), screen distance \(D_1 = 100\, cm\)
New slit distance \(d_2 = 2 - 0.5 = 1.5\, mm\), new screen distance \(D_2 = 100 + 50 = 150\, cm\)
Calculate new fringe width: \[ \beta_2 = \frac{\lambda D_2}{d_2} = \beta_1 \times \frac{D_2}{D_1} \times \frac{d_1}{d_2} = 0.36 \times \frac{150}{100} \times \frac{2}{1.5} = 0.36 \times 1.5 \times \frac{4}{3} = 0.36 \times 2 = 0.72\, mm \]